The equilibrium constant for decomposition of is at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation of water is (nearest integer value). [Assume is negligible with respect to 1]

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Solution

Q.
The equilibrium constant for decomposition of $H_{2} O (g)$
$H_{2} O(g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g) (Δ G ° = 92.34 {kJ mol}^{- 1})$

is $8.0 \times 10^{- 3}$ at 2300 K and total pressure at equilibrium is 1 bar.
Under this condition, the degree of dissociation $(α)$ of water is $_____\times 10^{- 2}$ (nearest integer value).

[Assume $α$ is negligible with respect to 1] [2025]

Ans.
(5)

$\begin{matrix} H_{2} O (g) & ⇌ & H_{2} (g) & + & \frac{1}{2} O_{2} (g) \\ t =0 (mol) & a & 0 & 0 \\ t =∞ (mol) & a (1 - α) & a α & \frac{a α}{2} & Total= a (1 + \frac{α}{2}) \\ \approx a (as α ≪ 1) \\ \approx a (as α ≪ 1) \\ t =∞ (mol fraction) & \frac{a}{a} = 1 & \frac{a α}{a} = α & \frac{a α}{2 a} = \frac{α}{2} \\ t =∞ & 1 \times P & α \times P & \frac{α}{2} \times P \\ (partial & = 1 \times 1 & = α \times 1 & = \frac{α}{2} \times 1 \\ pressure) & = 1 & = α & = \frac{α}{2} \end{matrix}$

$K_{p} = \frac{{(P_{O_{2}})}^{1 / 2} \times P_{H_{2}}}{P_{H_{2} O}} = \frac{{(\frac{α}{2})}^{1 / 2} \times α}{1} = \frac{α^{3 / 2}}{\sqrt{2}}$

$8 \times 10^{- 3} = \frac{α^{3 / 2}}{\sqrt{2}}$

$α^{3} = 128 \times 10^{- 6}$

$α = {(128)}^{1 / 3} \times 10^{- 2} = 5.03 \times 10^{- 2}$

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